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Examples Based on Hyperbola 

Illustration 1: If the chords of the hyperbola x2 – y2 = a2 touch the parabola y2 = 4ax, then the locus of the middle points of these chords is the curve

(a) y2 (x + a) = x3 

(b) y2 (x – a) = x3 

(c) y2 (x + 2a) = 3x3 

(d) y2 (x – 2a) = 2x3 

Solution: Let the mid-point of the chord be (h,k)

Then, the equation of the chord of x2 – y2  = a2 is

T = S1

hx – ky = h2 - k2

So, y = h/k x – (h2 - k2)/k and this is tangent of y2 = 4ax

so, – (h2 – k2)/k = a/(h/k)

or -(h2 - k2) = ak2/h

or -h3 + hk2 = ak2

or k2(h-a) = h3

Hence, locus of mid-point is y2(x-a) = x3.

 

Illustration 2: Find the equation of the hyperbola whose directrix is 2x + y = 1, focus is (1, 1) and eccentricity is √3.

Solution: Let S(1, 1) be focus and P(x, y) be any point on the hyperbola. From P draw PM perpendicular to the directrix then PM = (2x + y – 1)/√(22 + 12) = (2x + y – 1)/√5

Also from the definition of the hyperbola, we have

SP/PM = e ⇒ SP = ePM

⇒ √(x–1)2 + (y–1)2 = √3 (2x + y – 1/√5)

⇒ (x – 1)2 + (y – 1)2 = 3 (2x + y – 1)2/5

⇒ 5[(x2 + 1 – 2x) + (y2 + 1 – 2y)] = 3(4x2 + y2 + 1 + 4xy – 4x – 2y)

⇒ 7x2 – 2y2 + 12xy – 2x – 4y – 7 = 0

 

Illustration 3: The normal at P to a hyperbola of eccentricity e intersects its transverse and conjugate axes at L and M respectively. If locus of the mid-point of LM is hyperbola, then eccentricity of the hyperbola is 

(a) (e + 1)/(e - 1)

(b) e/√(e2 - 1)

(c) e

(d) none of these

Solution: Equation of normal at P (a sec φ, b tan φ) is

ax cos φ + by cot φ = a2 + b2

Hence, the coordinates of L and M are

L = [(a2 + b2)sec φ/a, 0 ]

and M = [0, (a2 + b2)tan φ/b] respectively

Let mid point of LM be Q(h,k)

then h = (a2 + b2)sec φ/2a

So, sec φ = 2ah/(a2+b2)

and k = (a2+b2)tan φ/2b

so, tan φ = 2bk/(a2+b2)

so, we have sec2 φ – tan2φ = 4a2h2/(a2+b2)2 – 4b2k2/(a2+b2)2

Hence, the locus

x2/[(a2+b2)/2a]2 – y2/[(a2+b2)/2b]2 = 1

Suppose eccentricity is e1.

then, [(a2+b2)/2b]2 = [(a2+b2)/2a]2 (e12 – 1)

So, a2 = b2 (e12 -1)

a2 = a2(e2-1)(e12-1)

So, e2e12 – e2  -e12 + 1 = 1

So, e12 (e2-1) = e2

e1 = e/√(e2-1)

 

Illustration 4: The equations to the common tangents to the two hyperbolas x2/a2 – y2/b2 = 1 and y2/a2 – x2/b2 = 1 are

(a) y = ± x ± √(b2 - a2)

(b) y = ± x ± √(a2 - b2)

(c) y = ± x ± (a2 – b2)

(d) y = ± x ± √(b2 + a2)

Solution: Let y = mx+c be a common tangent of hyperbola

x2/a2 – y2/b2 = 1 and

y2/a2 – x2/b2 = 1

The condition of tangency for the first equation is c2 = a2m2 – b2     ….. (1)

and the condition of tangency for the second equation is c2 = a2 - b2m2

From these two equations it follows that,

a2m2 – b2 = a2 – b2m2

So, a2(m2 – 1) + b2(m2 – 1) = 0

so, (a2 + b2)(m2 - 1) = 0

since, a2 + b2 ≠ 0, so m2 - 1 = 0

So, m = ± 1.

From eq.(1) c2 = a2 - b2

So, c = ± √(a2 – b2)

Hence, the equations of common tangents are y = ± x ± √(a2 – b2)

 

Illustration 5: Find the directrix, foci and eccentricity of the hyperbola ax2 – y2 = 1

Solution: The given hyperbola is ax2 – y2 = 1

or     x2/(1/a) – y2/1 = 1                                                         …… (1)

which is of the form (x2/a2) – (y2/b2) = 1

Here a2 = 1/a, b2 = 1

If e be the eccentricity of the hyperbola, then

b2 = a2(e2 – 1)

⇒ 1 =  (e2 – 1)

⇒ a = (e2 – 1)

or e2 = a + 1 or e = √(a + 1)

Also the foci are given by (+ ae, 0)

∴ The required foci are (+ 1/√a √(a+1), 0)

or (+ √a+1/a, 0)

And the directrix is given by x = + (a/e)

⇒ x = + [1/√a/√a+1]                    (since a = 1/√a, e = 1/√a+1)

x = + 1/√a(a+1)

 

Illustration 6: Find the locus of a point, the difference of whose distances from two fixed points is constant.

Solution: Let two fixed points be S (ae, 0) and S’ (–ae, 0).

Let P(x, y) be a moving point such that

SP – S’P = Constant = 2a (say).

Then √[(x – ae)2 + (y – 0)2] – √[x + ae]2 + (y – 0)2 = 2a

⇒ √[(x – ae)2 + y2] = + 2a + √[(x + ae)2 + y2

⇒ (x – ae)2 + y2 = 4a2 + (x + a2)2 + y2 + 4a √[(x – ae)2 + y2]

⇒ (x – ae)2 – (x + ae)2 – 4a2 = + 4a √[(x – ae)2 + y2]

⇒ –4aex – 4a2 = + 4a √[(x – ae)2 + y2]

⇒ (ex + a)2 = (x + ae)2 + y2

⇒ (e2 – 1)x2 – y2 = a2(e2 – 1)

⇒ x2/a2 – y2/b2 = 1 taking b2 = a2(e2 – 1)

This is a hyperbola.

 

Illustration 7: If A, B, C are three points on the hyperbola xy = c2 and AC is perpendicular to BC, prove that AB is parallel to the normal to the curve at C.

Solution: Let the three points A, B, C respectively be (ct1, c/t1), (ct2, c/t2) and (ct3, c/t3). Since AC is perpendicular to BC,

(c/t3–c/t1/ct3–ct1) = – 1 ⇒ t1t2 = –1                                 …… (1)

Normal to the curve at C (ct3, c/t3) is

y = xt32 + 2/t3 (1 – t34) and its slope is t32 = –1/t1t2              …… (2)

Slope of AB = (c/t2–c/t1/ct2–ct1) = – 1/t1t2 = t32

⇒ AB is parallel to the normal at C.

 

Illustration 8: Find the equation of the hyperbola the distance between whose foci is 16, whose eccentricity is √2 and whose axis is along the x-axis centre being the origin.

Solution: We have b2 = a2(e2 – 1) = a2 

⇒ b = a.

Also 2ae = 16 ⇒ ae = 8 ⇒ a = 4√2.

Hence the equation of the required hyperbola is

x2/32 – y2/32 = 1 ⇒ x2 – y2 = 32.

 

Illustration 9: The perpendiculars drawn from the centre of a hyperbola x2/a2 – y2/b2 = 1 upon the tangent and normal at any point of the hyperbola meet them in Q and R. Find the locus of Q and R.

Solution: Tangent at any point P(a sec θ, b tan θ) is  sec θ – y/b  tan θ = 1. … (1)

Equation of the line through centre (origin) perpendicular to (1) is y = –a sin θ/b x

⇒ sin θ = – by/ax

Eliminating θ from (1), we get x/a cos θ – y/b cos θ (–by/ax) = 1.

⇒ x2 + y2 = ax cos θ ⇒ (x2 + y2)2 = a2x2(1 – b2y2/a2x2)

Or (x2 + y2)2 = a2x2 – b2y2, which is the locus of Q.

Normal at the point P (a sec θ, b tan θ) is ax cos θ + by

cot θ = a2 + b2                                                         … (2)

Equation of the line perpendicular to (2) drawn from the centre is

y = bx/a sin θ                                                     … (3)

Form (2) and (3),

sin θ = bx/ay and ax √1–b2y2/a2x2 + by √1–b2x2/a2y2 . ay/bx = a2 + b2

⇒ (x2 + y2)2 (a2y2 – b2x2) = (a2 + b2)2 x2y2, which is the locus of R.

 

Illustration 10: A straight line touches the rectangular hyperbola 9x2 - 9y2  = 8 and the parabola y2 = 32x. The equation of the line is:

(a) 9x + 3y – 8 = 0

(b) 9x - 3y + 8 = 0

(c) 9x + 3y + 8 = 0

(d) 9x - 3y – 8 = 0

Solution: Equation of tangent in terms of slope of y2 = 32x is

y = mx + 8/m  …... (1)

which is also a tangent of 9x2 – 9y2 = 8

So, x2 – y2 = 8/9

Hence, (8/m)2 = 8/9 m2 – 8/9

So, 8/m2 = m2/9 – 1/9

so, 72 = m4 – m2

So, m4 – m2 – 72 = 0

(m2 – 9)(m2 + 8) = 0

So, m = 9 and m2 + 8 ≠ 0

So, m = ± 3

So, from eq (1), we have y = ± 3x ± 8/3

So, 3y = ± 9x ± 8

or ± 9x – 3y ± 8 = 0

i.e. 9x – 3y + 8 = 0, 9x – 3y – 8 = 0

 -9x – 3y + 8 = 0, – 9x – 3y – 8 = 0

or 9x – 3y + 8 = 0, 9x – 3y – 8 = 0

9x + 3y - 8 = 0, 9x + 3y + 8 = 0

 

Illustration 11: Assertion/Reason Based:

Choose the correct option out of the following four:

(a) Both A and R are individually true and R is the correct explanation of A.

(b) Both A and R are individually true but R is not the correct explanation of A.

(c) A is true but R is false

(d) A is false but R is true.

Assertion: Director circle of hyperbola x2/a2 – y2/b2 + 1 = 0 is defined only when b > a.

Reason: Director circle of hyperbola x2/25 – y2/9 = 1 is x2 + y2 = 16.

(a) A

(b) B

(c) C

(d) D

Solution: Since, the director circle is the locus of point of intersection of perpendicular tangents

Any tangent in terms of m of x2/a2 – y2/b2 + 1 = 0 is

y = mx ± √(b2 - a2m2)

or (y – mx)2 = b2 – a2m2

so, m2(x2 + a2) -2mxy + y2 - b2 = 0

since, m1m2 = -1

So, (y2 - b2)/(x2 + a2) = -1

or x2 + y2 = b2 - a2, b > a

Also, director circle of x2/25 – y2/9 = 1 is

x2 + y2 = 25 – 9 = 16.

 

Illustration 12: C is a centre of the hyperbola x2/a2 – y2/b2 = 1 and the tangent at any point P meets asymptotes in the point Q and R. Find the equation to locus of the centre of the circle circumscribing the triangle CQR.

Solution: This equation to the hyperbola is given as

x2/a2 – y2/b2 = 1                                      …… (1)

Let P any point on it as (a sec θ, b tan θ), then the equation of tangent at P is

x/a – y/b sin θ = cos θ                              …… (2)

The equation to the asymptotes to (1) are

x/a = y/b                                                …… (3)

and x/a = – y/b                                             …… (4)

Solving (2) and (3), we get the coordinates of Q as

(a cos θ/1–sin θ, bcosθ/1–sinθ)

Solving (2) and (4), we get the coordinates of R as

(acosθ/1+sinθ, –bcosθ/1+sinθ)

Let O be the centre of the circle passing through C, Q and R having its coordinates as (h, k). Then clearly OC = OQ

⇒ h2 + k2 = (h–acosθ/1–sinθ)2 + (k–bcosθ/1–sinθ)2

⇒ h2 + k2 = (h–acosθ/1–sinθ)2 + (k–bcosθ/1–sinθ)2

⇒ h2 + k2 = h2 + k2 + (a2 + b2) cos2θ/(1–sinθ)2 – (2ah + 2bk) cosθ/1–sinθ

⇒ 2(ah + bk) = (a2 + b2) cosθ/1–sinθ                              …… (5)

Similarly OC = OR

Hence h2 + k2 = (h–acosθ/1+sinθ)2 + (k+bcosθ/1+sinθ)2

Which on simplification as in the last case, given 2(ah –bk) = (a2 + b2) cosθ/1–sinθ                                   …… (6)

to get the locus of the point O we have to eliminate f from (5) and (6), so multiplying the two we get

4(a2h2 – b2k2) = (a2 + b2) cos2θ/1–sin2θ = (a2 + b2)

for (h, k), we get the required locus as

4(a2x2 – b2y2) = (a2 + b2)2

 

Illustration 13: The equation of a hyperbola conjugate to the hyperbola x2 + 3xy + 2y2 + 2x + 3y = 0 is

(a) x2 + 3xy + 2y2 + 2x + 3y + 1 = 0

(b) x2 + 3xy + 2y2 + 2x + 3y + 2 = 0

(c) x2 + 3xy + 2y2 + 2x + 3y + 3 = 0

(d) x2 + 3xy + 2y2 + 2x + 3y + 4 = 0

Solution: Since, the given hyperbola is H : x2 + 3xy + 2y2 + 2x + 3y = 0

Let the pair of asymptotes be x2 + 3xy + 2y2 + 2x + 3y + λ = 0

So, Δ = 0

1.2.λ + 2.3/2.1.3/2 – 1.9/4 – 2.1 - λ.9/4 = 0

-λ/4 + 9/2 - 9/4 – 2 = 0

λ/4 = 9/4 – 2 = ¼

So, λ = 1

Therefore, A : x2 + 3xy + 2y2 + 2x + 3y + 1 = 0

since, H + C = 2A

So, C = 2A – H

         = x2 + 3xy + 2y2 + 2x + 3y + 2

Hence, the conjugate hyperbola is x2 + 3xy + 2y2 + 2x + 3y + 2 = 0

 

Illustration 14: The locus of the point of intersection of two perpendicular tangents to the hyperbola x2/a2 – y2/b2 = 1 is

(a) director circle

(b) x2 + y2 = a2

(c) x2 + y2 = a2 - b2

(d) x2 + y2 = a2 + b2

Solution: Equation of any tangent in terms of slope m is

y = mx + (a2m2 – b2)

It passes through (h, k), so we have

(k - mh)2 = a2m2 – b2

So, m2(h2 - a2) – 2mhk + k2 + b2 = 0

This is a quadratic in m

Let the slopes of tangents be m1 and m2.

then m1m2 = -1.

So, (k2 + b2)/(h2 – a2) = -1

(h2 + k2) = (a2 – b2)

Hence, the locus is (x2 + y2) = (a2 – b2) which is the director circle of x2/a2 – y2/b2 = 1.

 

Illustration 15: If a tangent to the hyperbola x2/25 – y2/16 = 1 makes equal intercepts of lengths λ on the axes, then what is the value of λ?

Solution: Equation of a tangent at (5 sec θ, 4 tan θ) to the given hyperbola is x/5 sec θ – y/4 tan θ = 1

This intersects the axes at (5 cos θ, 0) and (0, – 4 cot θ).

So, we have  λ = |5 cos θ| = |4 cot θ|

So, λ2 = 25 cos2θ = 16 cot2θ

So, sec2θ = 25/ λ2 and tan2θ = 16/λ2

So, sec2θ – tan2θ = 25/λ2 – 16/λ2 =1

so, λ2 = 25 - 16 = 9

So, λ = 3.

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